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Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.
Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.
This Question has 1 answers.
Given that the third term of an arithmetic progression (A.P.) is 16:
$ a_3 = a + 2d = 16 \quad \cdots (1) $
It is also given that the seventh term exceeds the fifth term by 12:
$ a_7 = a + 6d $
$ a_5 = a + 4d $
From the given condition:
$ (a + 6d) - (a + 4d) = 12 $
$ 2d = 12 $
$ d = 6 $
Substituting $ d = 6 $ in equation (1):
$ a + 2(6) = 16 $
$ a + 12 = 16 $
$ a = 4 $
Thus, the A.P. has first term $ a = 4 $ and common difference $ d = 6 $.
Now, finding the sum of the first 29 terms:
The sum of the first $ n $ terms of an A.P. is given by:
$ S_n = \frac{n}{2} (2a + (n-1)d) $
Substituting $ n = 29 $, $ a = 4 $, and $ d = 6 $:
$ S_{29} = \frac{29}{2} (2(4) + (29-1)(6)) $
$ = \frac{29}{2} (8 + 168) $
$ = \frac{29}{2} \times 176 $
$ = 29 \times 88 $
$ = 2552 $
Thus, the required A.P. is $ 4, 10, 16, 22, \dots $ and the sum of the first 29 terms is $ 2552 $.
$ a_3 = a + 2d = 16 \quad \cdots (1) $
It is also given that the seventh term exceeds the fifth term by 12:
$ a_7 = a + 6d $
$ a_5 = a + 4d $
From the given condition:
$ (a + 6d) - (a + 4d) = 12 $
$ 2d = 12 $
$ d = 6 $
Substituting $ d = 6 $ in equation (1):
$ a + 2(6) = 16 $
$ a + 12 = 16 $
$ a = 4 $
Thus, the A.P. has first term $ a = 4 $ and common difference $ d = 6 $.
Now, finding the sum of the first 29 terms:
The sum of the first $ n $ terms of an A.P. is given by:
$ S_n = \frac{n}{2} (2a + (n-1)d) $
Substituting $ n = 29 $, $ a = 4 $, and $ d = 6 $:
$ S_{29} = \frac{29}{2} (2(4) + (29-1)(6)) $
$ = \frac{29}{2} (8 + 168) $
$ = \frac{29}{2} \times 176 $
$ = 29 \times 88 $
$ = 2552 $
Thus, the required A.P. is $ 4, 10, 16, 22, \dots $ and the sum of the first 29 terms is $ 2552 $.
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